Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Review - Page 598: 103


$x=\dfrac{\log\dfrac{1}{2}+\log5}{\log5}\\\\ x\approx0.5693 $

Work Step by Step

Getting the logarithm of both sides and then using the properties of logarithms, the value of $x$ in the expression $ 5^{x-1}=\dfrac{1}{2} $ is \begin{array}{l} \log5^{x-1}=\log\dfrac{1}{2}\\\\ (x-1)\log5=\log\dfrac{1}{2}\\\\ x\log5-\log5=\log\dfrac{1}{2}\\\\ x\log5=\log\dfrac{1}{2}+\log5\\\\ x=\dfrac{\log\dfrac{1}{2}+\log5}{\log5}\\\\ x\approx0.5693 .\end{array}
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